Question

Two infinitely long straight wires $A$ and $B$, each carrying current $I$ are placed on $x$ and $y$ -axis, respectively. The current in wires $A$ and $B$ flow along $-\hat{i}$ and $\hat{j}$ directions respectively. The force on a charged particle having charge $q$, moving from position, $r=d(\hat{i}+\hat{j})$ with velocity $v=v\hat{i}$ is

• A. $\frac{{\mu }_{0}lqv}{2\pi d}\hat{j}$
• B. $\frac{{\mu }_{0}lqv}{\pi d}\hat{j}$
• C. $\frac{{\mu }_{0}lqv}{\sqrt{2}\pi d}\hat{k}$
• D. 0

Answer 1

Answer: (B)

The given situation for two current carrying wires are shown in the figures.

Position vector of charge particle,
$r=d(\hat{i}+\hat{j})=d\hat{i}+\widehat{dj}$
The magnetic field intensity due to the wire along $-\hat{i}$ direction,
$B_1=\frac{{\mu }_0}{2\pi }\cdot \frac{I}{d}(-\hat{k})$
Similarly, magnetic field intensity due to wire along $\hat{j}$ direction.
$B_2=\frac{{\mu }_0}{2\pi }\cdot \frac{I}{d}(-\hat{k})$
$\therefore$ Resultant magnetic field at point $P$ due to both the wires,
$B=B_1+B_2=2\cdot \frac{{\mu }_0}{2\pi }\cdot \frac{I}{d}(-\hat{k})$
$B=\frac{{\mu }_0}{\pi }\cdot \frac{I}{d}(-\hat{k})$
$\therefore$ Force on the charged particles,
$F=q(v\times B)=q\left[v\hat{i}\times \frac{{\mu }_0}{\pi }\frac{I}{d}\cdot (-\hat{k})\right]=\frac{{\mu }_{0}Iqv}{\pi d}\cdot \hat{j}$